package com.zyjblogs.lock;

/**
 * 测试 ：生成者消费者模型->利用缓冲区解决：管程法
 */
public class TestPC {
    public static void main(String[] args) {
        SynContainer container = new SynContainer();
        new Productor(container).start();
        new Consumer(container).start();
    }
}

class Productor extends Thread {
    SynContainer container;

    public Productor(SynContainer container) {
        this.container = container;
    }
    //生产

    @Override
    public void run() {
        for (int i = 0; i < 100; i++) {
            container.push(new Ckicken(i));
            System.out.println("生产了" + i + "只鸡");
        }
    }
}

//消费者
class Consumer extends Thread {
    SynContainer container;

    public Consumer(SynContainer container) {
        this.container = container;
    }
    //消费

    @Override
    public void run() {
        for (int i = 0; i < 100; i++) {
            System.out.println("消费了-->" + container.pop().id +"只鸡");
        }
    }
}

//产品
class Ckicken {
    int id;
    Ckicken(int id) {
        this.id = id;
    }
}

class SynContainer {
    //需要一个容器大小
    Ckicken[] ckickens = new Ckicken[10];
    //容器计算器
    int count = 0;

    //生产者放入产品
    public synchronized void push(Ckicken ckicken) {
        //如果容器满了，就需要等待消费者消费
        if (count == ckickens.length) {
            try {
                this.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            //通知消费者消费
        }
        //如果没有慢，我们就需要丢入产品
        ckickens[count] = ckicken;
        count++;
        this.notifyAll();
    }

    //消费者消费产品
    public synchronized Ckicken pop() {
        if (count == 0) {
            //等待生产者生产，消费者等待
            try {
                this.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        //如果可以消费
        count--;
        this.notifyAll();
        //吃完了，通知生产者生产
        return ckickens[count];

    }
}